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<h1>Bridge Position</h1>

<p>In this section we will develop the formula for the bridge Position BP.</p>

<p class="imgbox"><a name="fig-058"></a><img src="../img/fig-058.png" style="max-width: 400px;"/><br />Figure 58: Calculating the bridge positon.</p>

<h2>What We Know</h2>

<pre class="formula">
(1)	P0P1 = P0P2 = P0P3 = P0P6 = DR = dome radius
(2)	P1P2 = BL = body length
(3)	P5P6 = DH = dome height at bridge position</li>
(4)	P5P6 is perpendicular to P1P2
</pre>

<h2>What We Want</h2>

<pre class="formula">
(5)	P1P5 = BP = bridge position
</pre>

<h2>Calculation Path</h2>

<p>We start with drawing a helper line P0-P6, which is the dome radius line through P6 at the bridge position P5. </p>

<p>Then we drop a perpendicular bisector of P1-P0-P2 from P0 through P4 and P3.</p>

<p class="imgbox"><a name="fig-046"></a><img src="../img/fig-046.png" style="max-width: 400px;"/><br />Figure 46</p>

<p>Because the sides P0-P1 and P0-P6 are of equal length, P4 divides P1-P2 into halves:</p>

<pre class="formula">
(6)	P1P4 = P4P2 = P1P2 / 2
</pre>

<p>This creates a triangle P0-P4-P1 with a right angle at P4. The length of the hypothenusis and the side P-P4 is known, so we can calculate the length of the 3rd side by applying the pythagorean theorem.</p>

<pre class="formula">
(7)	P0P4<sup>2</sup> + P1P4<sup>2</sup> = P0P1<sup>2</sup>
(8)	P0P4<sup>2</sup> = P0P1<sup>2</sup> &minus;  P1P4<sup>2</sup>
(9)	P0P4 = sqrt( P0P1<sup>2</sup> &minus;  P1P4<sup>2</sup> )
</pre>

<p>Then we draw another helper line, parallele to the reference line P1-P2 through P6. The intersection with P0P3 will be named P7. This creates a triangle P0-P7-P6. The angle at P7 is square, because both P0-P3 and P5-P6 are perpendicular to the reference line P1-P2 (equation (4)), and P6-P7 is parallel to P1-P2. Essentially, P4-P5-P6-P7 form a rectangle, so we can conclude:</p>

<pre class="formula">
(10)	P4P7 = P5P6
(11)	P4P5 = P6P7
</pre>

<p>The length of the hypothenusis P0-P6 of this triangle is known from equation (1), and the side P0-P7 can be can be calculated:</p>

<pre class="formula">
(12)	P0P7 = P0P4 + P4P7 = P0P4 + P5P6
</pre>

<p>Now we can use Pythagoras to calculate the distance of the bridge from the center point P4:</p>

<pre class="formula">
(13)	P6P7<sup>2</sup> + P0P7<sup>2</sup> = P0P6<sup>2</sup>
(14)	P6P7 = sqrt( P0P6<sup>2</sup> &minus; P0P7<sup>2</sup> )
</pre>

<p>The bridge position P1-P5 can be expressed as sum of parts, and with (11) we can write:</p>

<pre class="formula">
(15)	P1P5 = P1P4 + P4P5
</pre>

<p>With equation (11) we can write the final formula for the bridge position:</p>

<pre class="formula intermediate">
(16)	P1P5 = P1P4 + P6P7
</pre>

<h2>Substitution of Intermediate Results</h2>

<p>Introducing (14) into (16) :</p>

<pre class="formula">
(17)	P1P5 = P1P4 + sqrt( P0P6<sup>2</sup> &minus; P0P7<sup>2</sup> )
</pre>

<p>Substituting P0P7 from equation (12):</p>

<pre class="formula">
(18)	P1P5 = P1P4 + sqrt( P0P6<sup>2</sup> &minus; (P0P4 + P5P6)<sup>2</sup> )
</pre>

<p>Substituting P0P4 from equation (9):</p>

<pre class="formula">
(19)	P1P5 = P1P4 + sqrt( P0P6<sup>2</sup> &minus; (sqrt( P0P1<sup>2</sup> &minus;  P1P4<sup>2</sup> ) + P5P6)<sup>2</sup> )
</pre>

<p>Substituting P1P4 from equation (6):</p>

<pre class="formula">
(20)	P1P5 = P1P2 / 2 + sqrt( P0P6<sup>2</sup> &minus; (sqrt( P0P1<sup>2</sup> &minus;  (P1P2 / 2)<sup>2</sup> ) + P5P6)<sup>2</sup> )
</pre>

<p>The last step replaces the generic values with the names defined in equations (1) to (5), which gives us the final formla:</p>

<pre class="formula final">
(21)	BP = BL / 2 + sqrt( DR<sup>2</sup> &minus; (sqrt( DR<sup>2</sup> &minus;  (BL / 2)<sup>2</sup> ) + DH)<sup>2</sup> )
</pre>

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